牛客国庆集训派对Day6 A Birthday 费用流

牛客国庆集训派对Day6 A Birthday：https://www.nowcoder.com/acm/contest/206/A

题意：

恬恬的生日临近了。宇扬给她准备了一个蛋糕。
正如往常一样，宇扬在蛋糕上插了n支蜡烛，并把蛋糕分为m个区域。因为某种原因，他必须把第i根蜡烛插在第a_i个区域或第b_i个区域。区域之间是不相交的。宇扬在一个区域内同时摆放x支蜡烛就要花费x²的时间。宇扬布置蛋糕所用的总时间是他在每个区域花的时间的和。
宇扬想快些见到恬恬，你能告诉他布置蛋糕最少需要多少时间吗？

 

思路:

　建立图，左边n个点，表示不同的蜡烛，右边m个点，表示不同的区域，根据ai和bi从左边向右边连容量为1，费用为0的边，右边每个m点都向汇点连n条边，容量为1，费用为多一个蜡烛的增量。左边还要从源点拉来容量为1，费用为0的边。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <iomanip>
#include   <cstdlib>
#include    <cstdio>
#include    <string>
#include    <vector>
#include    <bitset>
#include    <cctype>
#include     <queue>
#include     <cmath>
#include      <list>
#include       <map>
#include       <set>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int ,pii> p3;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFFLL;  //
const ll nmos = 0x80000000LL;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //
const double PI=acos(-1.0);

template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------show time----------------------*/

        const int maxn = 1e6+;

        struct Edge
        {
            int to,val,cost,nxt;
        }gEdge[maxn];
        int gHead[maxn],gPre[maxn];
        int gPath[maxn],gDist[maxn];
        bool in[maxn];
        int gcount = ;
        int n,m,k,w;
        int fac[maxn];
        bool spfa(int s,int t){

            memset(gPre, -, sizeof(gPre));
            memset(gDist,inf,sizeof(gDist));
            memset(in, false , sizeof(in));
            gDist[s] = ; in[s] = true;
            queue<int>q;
            q.push(s);
            while(!q.empty()){
                int u = q.front();
                q.pop();    in[u] = false;
                for(int e = gHead[u]; e!=-; e = gEdge[e].nxt){
                    int v = gEdge[e].to, w = gEdge[e].cost;
                    if(gEdge[e].val >  && gDist[v] > gDist[u] + w){
                        gDist[v] = gDist[u] + gEdge[e].cost;
                        gPre[v] = u;
                        gPath[v] = e;
                        if(!in[v]){
                            q.push(v);in[v] = true;
                        }
                    }
                }
            }
            if(gPre[t] == -)return false;
            return true;
        }
        int MinCostFlow(int s,int t){
            int cost = ,flow = ;
            while(spfa(s,t)){
                int f = inf;
                for(int u = t; u != s; u = gPre[u]){
                    if(gEdge[gPath[u]].val < f){
                        f =gEdge[gPath[u]].val;
                    }
                }
                flow += f;
                cost += gDist[t] * f;
                for(int u=t; u!=s; u = gPre[u]){
                    gEdge[gPath[u]].val -= f;
                    gEdge[gPath[u] ^ ].val += f;
                }
            }
            return cost;
        }

        void addedge(int u,int v,int val, int cost){
            gEdge[gcount].to = v;
            gEdge[gcount].val = val;
            gEdge[gcount].cost = cost;
            gEdge[gcount].nxt = gHead[u];
            gHead[u] = gcount++;

            gEdge[gcount].to = u;
            gEdge[gcount].val = ;
            gEdge[gcount].cost = -cost;
            gEdge[gcount].nxt = gHead[v];
            gHead[v] = gcount++;
        }

        struct eee
        {
            int l,r,w,op;
        }e[maxn];
        /*
是大源点，m+m+1是ci源点，m+m+2是终点。
        */
        void solve(){
                    memset(gHead,-,sizeof(gHead));
                    gcount = ;
                    scanf("%d%d", &n, &m);
                    for(int i=; i<=n; i++) fac[i] = i*i;
                    for(int i=n; i>=; i--) fac[i] = fac[i] - fac[i-];

                    for(int i=; i<=n; i++){
                        addedge(,i,,);
                    }

                    for(int i=; i<=n; i++){
                        int le,ri;
                        scanf("%d%d", &le, &ri);
                        addedge(i,le + n, , );
                        addedge(i,ri + n, , );
                    }

                    for(int i=; i<=m; i++){
                        for(int j=; j<=n; j++){
                            addedge(i+n,n+m+,,fac[j]);
                        }
                    }

                    printf("%d\n",MinCostFlow(,n+m+));
        }

int main(){
            solve();
            return ;
}